[QuantEcon]MATLAB混编FORTRAN语言

Nevertheless, we still illustrate with a typical example in linear model in the following for the purpose of comparison. Given $X, Y$ and $\lambda_0 > 0$, we want to find $\beta$ such that
$$
\min_{\beta} ||Y - X\beta||_2^2 + \lambda_0 ||\beta||_1,
$$
where, say, $\lambda_0 = 8$.

Suppose the `glmnet` fitted linear predictor at $\lambda$ is
$\hat\eta_\lambda(x)$ and the relaxed version is $\tilde
\eta_\lambda(x)$. We also allow for shrinkage between the two:
$$\tilde \eta_{\lambda,\gamma}=(1-\gamma)\tilde
\eta_\lambda(x)+\gamma\hat\eta_\lambda(x).$$
$\gamma\in[0,1]$ is an additional tuning parameter which can be
selected by cross validation.

$$\textbf{y}_i \sim \mathcal{N}(\mu_i, \sigma^2_1)$$
$$\mu_i = \alpha + \beta[\textbf{X}_{i,1}] + \gamma[\textbf{X}_{i,2}] + \delta \textbf{X}_{i,2}, \quad i=1,\dots,N$$
$$\epsilon_i = \textbf{y}_i - \mu_i$$
$$\alpha \sim \mathcal{N}(0, 1000)$$
$$\beta_j \sim \mathcal{N}(0, \sigma^2_2), \quad j=1,\dots,J$$
$$\beta_J = - \sum^{J-1}_{j=1} \beta_j$$
$$\gamma_k \sim \mathcal{N}(0, \sigma^2_3), \quad k=1,\dots,K$$
$$\gamma_K = - \sum^{K-1}_{k=1} \gamma_k$$
$$\delta \sim \mathcal{N}(0, 1000)$$
$$\sigma_m \sim \mathcal{HC}(25), \quad m=1,\dots,3$$

$$\textbf{y} \sim \mathcal{N}(\mu, \sigma^2_1)$$
$$\mu_i = \alpha + \beta[\textbf{x}_i], \quad i=1,\dots,N$$
$$\alpha \sim \mathcal{N}(0, 1000)$$
$$\beta_j \sim \mathcal{N}(0, \sigma^2_2), \quad j=1,\dots,J$$
$$\beta_J = - \displaystyle\sum^{J-1}_{j=1} \beta_j$$
$$\sigma_{1:2} \sim \mathcal{HC}(25)$$

$$\textbf{y}_i \sim \mathcal{N}(\mu_i, \sigma^2_1)$$
$$\mu_i = \alpha + \beta[\textbf{X}_{i,1}] + \gamma[\textbf{X}_{i,2}], \quad i=1,\dots,N$$
$$\epsilon_i = \textbf{y}_i - \mu_i$$
$$\alpha \sim \mathcal{N}(0, 1000)$$
$$\beta_j \sim \mathcal{N}(0, \sigma^2_2), \quad j=1,\dots,J$$
$$\beta_J = - \sum^{J-1}_{j=1} \beta_j$$
$$\gamma_k \sim \mathcal{N}(0, \sigma^2_3), \quad k=1,\dots,K$$
$$\gamma_K = - \sum^{K-1}_{k=1} \gamma_k$$
$$\sigma_m \sim \mathcal{HC}(25), \quad m=1,\dots,3$$

$$\textbf{y} = \mu + \epsilon$$
$$\mu = \textbf{X} \beta$$
$$\beta_j \sim \mathcal{N}(0, 1000), \quad j=1,\dots,J$$

\subsection{Form}
$$\textbf{y}_t \sim \mathcal{N}(\mu_t, \sigma^2_t), \quad t=1,\dots,T$$
$$\mu_t = \alpha + \sum^P_{p=1} \phi_p \textbf{y}_{t-p}, \quad t=1,\dots,T$$
$$\epsilon_t = \textbf{y}_t - \mu_t$$
$$\alpha \sim \mathcal{N}(0, 1000)$$
$$\phi_p \sim \mathcal{N}(0, 1000), \quad p=1,\dots,P$$
$$\sigma^2_t = \omega + \sum^Q_{q=1} \theta_q \epsilon^2_{t-q}, \quad t=2,\dots,T$$
$$\omega \sim \mathcal{HC}(25)$$
$$\theta_q \sim \mathcal{U}(0, 1), \quad q=1,\dots,Q$$

$$\Pr(A | B) = \frac{\Pr(B | A)\Pr(A)}{\Pr(B)}$$

For example, suppose one asks the question: what is the probability of going to Hell, conditional on consorting (or given that a person consorts) with Laplace's Demon\footnote{This example is, of course, intended with humor.}. By replacing $A$ with $Hell$ and $B$ with $Consort$, the question becomes

$$\Pr(\mathrm{Hell} | \mathrm{Consort}) = \frac{\Pr(\mathrm{Consort} | \mathrm{Hell}) \Pr(\mathrm{Hell})}{\Pr(\mathrm{Consort})}$$

Note that a common fallacy is to assume that $\Pr(A | B) = \Pr(B | A)$, which is called the conditional probability fallacy.

\subsection{Bayes' Theorem, Example 2} \label{bayestheorem2}

Another way to state Bayes' theorem is

$$\Pr(A_i | B) = \frac{\Pr(B | A_i)\Pr(A_i)}{\Pr(B | A_i)\Pr(A_i) +...+ \Pr(B | A_n)\Pr(A_n)}$$

\item $\Pr(A_1) = \Pr(\mathrm{Hell})$
\item $\Pr(A_2) = \Pr(\mathrm{Heaven})$
\item $\Pr(B) = \Pr(\mathrm{Consort})$
\item $\Pr(A_1 | B) = \Pr(\mathrm{Hell} | \mathrm{Consort})$
\item $\Pr(A_2 | B) = \Pr(\mathrm{Heaven} | \mathrm{Consort})$
\item $\Pr(B | A_1) = \Pr(\mathrm{Consort} | \mathrm{Hell})$
\item $\Pr(B | A_2) = \Pr(\mathrm{Consort} | \mathrm{Heaven})$


Laplace's Demon was conjured and asked for some data. He was glad to oblige.